text:
Why would a glider have water ballast? If it is trying to stay aloft without an engine, wouldn't it be better to be as light as possible?
So I was looking at the description of a ASW 27 B glider and ran across this statement:

Two water tanks in the wing plus a further 35 liter tank in the
  fuselage enable the ASW 27 B to carry more water ballast than any
  other 15 m glider and also give it the widest range of wing loadings

If a glider is trying to stay aloft as long as possible, wouldn't it be better to be light?  Why would you add ballast and be able to dump it?

Apart from all the speed benefits,  dumping ballast shortly before landing makes for great photos.  Occasionally it has also been used to irritate competitors (^_-)

@yankeekilo: Haha, good points!

You are also trying to fly far and fast, which requires energy. And weight+gravity is a great way to store energy.

Mass doesn't affect the maximum distance, only the maximum endurance.
For example, image two identical planes A and B: A weights 50kg less than B. Assuming no wind (horizontal / vertical) and speed of best glide, both gliders will land at the exact same spot.
The lighter airplane A however will arrive later than B, as the speed of best glide is less than for B. In conclusion you can say, that additional mass only increases cruise speed, but not the travel distance.
Glider competitions are most of the time a route you have to fly in the shortest time possible. So that means, if you have a higher speed of best glide, you can fly faster in competitions.
The only downside to having a higher weight is, that your liftrate in thermals will be decreased and due to the higher speed it is harder to center the thermals.
It is to some extend also possible to shift the Centre of Gravity (CG) with the added load. The further it is to the aft limit, the higher your maximum distance is. This is because you will have less down-force from the stabilizer required. (If the CG is at the front limit, you will need to pull the control stick in order to fly level, therefore you have more drag). However I think this is rather a positive side effect and most of the time the water is used for flying faster.
Source: I am a glider pilot and currently doing my ATPL training.

Another downside/caveat of higher loading is higher stall speed.

Yes, I kind of implied that with "higher speed it is harder to center the thermals", but you are absolutely right. Most of the time you center thermals just above the stall speed.

There are two factors: higher stall speed and higher sink ratio. Both have adverse effects on circling, esp. in weak/narrow thermals. But I mentioned higher stall speed mainly because of the additional care you have to take e.g. near terrain (ridges) or in case of a sudden landing which may not leave time to dump properly.

This is a good answer, but, about the CG, most modern sailplane have a specific and different ballast, located in the tailfin, not in the wings.

Actually, the heavier glider will fly farther from the same height. The higher wing loading lets it fly at a higher Reynolds number, and for a glider this translates into less friction drag.

@PeterKämpf Do you have a source for that?

This answers the question, but it doesn't address the aerodynamics: why does a heavier glider have a great max L/D speed?

One part of my answer isn't addressed in this answer:  Why would you want to dump the ballast, considering the advantages you list, wouldn't a fixed ballast be better?

@PeterKampf why does friction drag decrease with increased airspeed? I would have thought the opposite was true.

@pericynthion: You are right, and I meant to say drag coefficient. I need to rephrase my comment.

@Force: Reynolds number increases with speed, and the friction drag coefficient goes down. Flying at a higher wing loading means flying at a higher speed regime and at a lower friction drag coefficient, thus better L/D than at the same polar point at lower speed. This is basic aerodynamics - what source do you need?

@PeterKämpf Ah, that makes sense now. Thanks, I was worried I had a basic misunderstanding :)

@PeterKämpf As pointed out in the answer, I assume that both gliders are flying at the speed of best glide

@Force … yes, which means that one flies faster because it has a higher wing loading. You say as much in your answer. If one arrives earlier at the same spot, it must have flown faster.

@Lnafziger: Because the landing will be softer if you get rid of the extra weight/mass beforehand.

Why then do commercial airliners try to stay as light as possible? Is it purely for the fuel consumption when rising? You're saying that, once at cruise altitude, a plane with 10000kg extra weight would fly just as far as one without?

In addition to the other answers, let´s look at this L/D(=E) diagram of the enticing DG-1000 from DG Flugzeugbau (but fear not, 'tis true for all gliders) :

The best L/D ratio is equal for different wing loadings, but is occuring at different speeds - the higher the load, the higher speed. You can also see that the minimum/stall speed is also higher for higher loads.
The next diagram shows the polar curve:

You can see that the minimum sink rate occurs at lightest load. The heavier the load, the longer you will have to circle in the same thermal for a given height gain.
The loading is a tradeoff between higher average speed and less efficient climbing. In case of strong thermals and/or long glide intervals, the optimum moves toward more, in weak conditions towards less or no ballast. The good thing is that you can dump water rather quickly (also partially), so that in a competition you usually tend to fill up (and dump  in case) rather than start light (the Quintus e.g can take up to 250 liters!)
Aft ballast in the vertical tailplane is sometimes used to balance a forward CG caused by water in the wings -  depending on your ship, partial dumping can be problematic.
Of course there are many philosophies and tactical debates concerning the "water or no water" dispute, but once you´ve overtaken an identical, lighter ship with full wings and no height loss, you get to see how much fun ballast can be (until the next thermal, that is).

I'm tuning in more than 3 years late because I'm not fully satisfied with the answers here. Yes, Lnafziger, when you want to stay up as long as possible, the plane should be as light as possible. But sometimes you need to get down fast: This is when water ballast is added.
Force is right: Water ballast speeds everything up. But there is more to it.
Also StallSpin has a good point: Higher wing loading equals less disturbance by gusts.
But there are two points which should be considered as well:

Higher speed means higher Reynolds number. Since this number shows the ratio of inertial to viscous forces, it means that friction drag is relatively lower. The consequence is that the glider with the higher wing loading really flies a little further than the light glider when both fly at their best L/D speed. The difference is not huge but gives the heavier ship another speed advantage when it can leave the last thermal one turn earlier than the lighter glider.
But the higher Reynolds number makes an even bigger difference at low speed: Roll control is much improved with water ballast. At the Reynolds number range typical for the outer wing of a glider at low speed (much less than one million) the speed increase improves stall resistance and control power markedly.

Friction drag coefficient of a flat plate over Reynolds number (picture source). The curve for a glider is between the fully laminar and the fully turbulent ones. Note the double logarithmic axes.

Tactics: Water ballast is used mostly in competitions, and when several aircraft share one thermal, every pilot waits for the others to fly off to the next thermal. Watching the others tells her/him where the best route for minimum altitude loss is. This even makes the highest pilots in the thermal open their speed brakes, just to avoid leaving the thermal first. With water ballast the climb speed is reduced (higher sink plus bigger turning radius conspire to reduce the climb rate of the glider significantly), so the pilot with water ballast will even have a tactical advantage in the climb phase by flying a heavier ship.

@ymb1: No, heavier planes fly faster and descent faster, too. It does not make sense to add ballast in a powered airplane unless you need to correct the cg location.

Although it is very clear that you are speaking of a general "need for speed", your specific comment "But sometimes you need to get down fast: This is when water ballast is added." seems to have caused some confusion.   (See for example https://aviation.stackexchange.com/q/75181/34686 ).  You might possibly consider changing to something to the effect of "But sometimes you need to fly fast.  This is when water ballast is added."

Force's answer is very good.  But the statement "additional mass only increases cruise speed, but not the travel distance", true for any one glide, doesn't take account of the fact that conditions suitable for soaring typically exist for a limited time each day - so increasing cruise speed definitely does increase distance.
StallSpin's point about reduced effect of turbulence on a ballasted glider is significant.  This is best seen when flying a ridge, which in strong wind can be very rough.  The ballasted glider, suffering less acceleration imposed by the rough air, can fly faster and lower, where the horizontal wind component is less, requiring a smaller crab angle.

Force's answer is pretty much the answer, but also consider that mass = inertia. If you weigh more, you are less likely to be disturbed by any given outside force (turbulence). A lighter plane is more maneuverable but it will also bounce around a lot.
I cannot comment on how much of an effect the ballasts in question have on this for a glider, though.

Probably a lot, given that gliders are generally much much lighter than internally-powered aircraft of the same size.

Another factor the existing answers don't mention: if you are flying a two-seater glider alone, you might want to add ballast to correct your center of gravity.
Gliders are light, so a missing person can have a significant effect on the center of gravity. Two-seaters are optimized for flying with two people aboard. I've even seen lead ballast being used in the nose of a glider when a very thin and small trainee was flying with a heavy-set instructor in the back seat.

That's why I own over 50 pounds of lead shot, sewed up into cloth bags.  I sit on it while flying a glider.  Other, heavier, pilots sit on something else.  Actually, in all seriousness, this answer could be improved a bit by expanding it to cover single-seat gliders.  In virtually all single-seat gliders the pilot sits well ahead of the CG, and so a light pilot very often must carry ballast at the pilot location to keep the CG within the allowable envelope.  Not really what the original question was asking about though.

In addition to all the good content in all the other good answers, one more point should be made: when the airmass is moving horizontally and/ or vertically, the glide ratio over the ground is different than the glide ratio through the airmass, and therefore the glide ratio over the ground is different than the L/D ratio.
When gliding into a headwind, the maximum obtainable glide ratio relative to the ground is higher when the glider is heavy than when it is light.  You easily verify this for yourself:  starting with the second diagram in this related answer, extend the horizontal axis left far enough to include the origin of the graph.  Now put your pencil on the point (x=50 kph, y=0).  Starting from this point (x= 50 kph, y=0), the slope of a line drawn tangent to the airspeed-versus-sink-rate curve is the highest obtainable glide ratio in a 50 kph headwind in air that is neither rising nor sinking.  You can see that the line drawn tangent to the ballasted curve is flatter (i.e. has less slope) than the line drawn tangent to the unballasted curve.
When we consider that when a glider flies a task that returns to the starting point on a windy day, it invariably spends more time flying with a headwind component than with a tailwind component, this is not a trivial point.
Naturally, this effect is even more pronounced if we draw our tangent line from (x=100 kph, y=0), representing the best achievable glide ratio when flying against a 100 kph headwind.
When slope-soaring a radio-controlled miniature glider in strong wind, it is not uncommon to encounter conditions where a lightly-loaded glider has difficulty making any forward progress at all and just sinks almost vertically down to the ground, while a heavily-loaded version of the same aircraft can be flown much closer to the max L/D angle-of-attack and thus can race forward at high speed while maintaining altitude or climbing.
Similarly, if we take the graph discussed above and extend the y axis upward so that extends into positive values for y, and start drawing our tangent line from the point (x=0, y=.2 m/s), we can find the highest obtainable glide ratio relative to the ground in the presence of a .2 m/s downdraft and zero headwind/ tailwind.  Again the line drawn tangent to the ballasted curve is flatter (i.e. has less slope) than the line drawn tangent to the unballasted curve.  In a downdraft, the maximum obtainable glide ratio relative to the ground is higher when the glider is heavy than when it is light.  Since the air between thermals is often sinking to some degree, this is not a trivial point either.  One instance where a glider pilot is most likely to be interested in maximizing his or her glide ratio over the ground is when he or she is flying in sinking air, and in this situation ballast helps.
The same method can used to find the maximum obtainable glide ratio relative to the ground in air that is sinking and includes a headwind component.  In this case ballast really helps a lot-- the maximum obtainable glide ratio relative to the ground will be much higher in the ballasted glider than in the unballasted glider.

Obviously this answer could be improved by actually providing figures with the relevant lines drawn on them, or links to another website that independently illustrates the same process-- I'll save that for some other day.

One point not explicitly mentioned in other answers is that on a strong soaring day, no one flies at best L/D.  Suppose the lift is strong and climbing is no problem.  Ballast up to max gross. Cruise between thermals at 100 knots. The sink rate at the same airspeed will be much higher if no ballast is carried.
You can easily see this by checking the polar diagrams provided in this related answer.  Look at the second diagram-- the graph of sink rate versus airspeed for three different wing loadings.  100 knots is about 180 kph.  At the heaviest loading, the sink rate at this airspeed is 1.8 m/sec, and at the lightest loading the sink rate at this airspeed is 3 m/sec.  That is a 66% higher sink rate.
When flying at some given airspeed that is well above best L/D speed, ballast does in fact increase the distance made good for the same loss of altitude.

"Check the ploar diagram for 100 knots at max gross", that would be nice if you could include this document, because that's a strange thing that the range can be increased just by adding weight... if it was true, commercial aircraft would carry more passenger on a greater distance for a lower cost.

For me to believe this answer you'll need a very good source to back it up.

Wow, lots of downvotes for this very true statement? Guess downvotes are a badge of accuracy on this site.  No wonder he gave up after just one answer.  Plus it is obvious the statement is true just by looking at the second diagram on this related answer https://aviation.stackexchange.com/a/698/34686 .

@mins -- the last line of your comment above is addressed by the last two paragraphs of my related answer here https://aviation.stackexchange.com/a/75217/34686

@quietflyer: "Wow, lots of downvotes for this very true statement?", the downvotes are likely not because this is true or false, but because the strange point (the heavier the aircraft, the longer the range) is not demonstrated and therefore "the answer is not useful" which is the current meaning of a downvote (it was different in the past I think). Compare with Peter's answer which is argued. Maybe you can improve the post.

@mins -- actually, he demonstrated it pretty well; he just didn't make it clear exactly which graph he was talking about, which was one provided in another recent answer.  Should be clear now.

@Notts90supportsMonica -- source now provided.


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What are the effects of headwind strength on landing?
Assuming a direct headwind with zero crosswind component, what are the effects of landing into say a 10kt vs 30kt vs 50kt headwind? Are there any specific advantages or drawbacks to a stronger headwind when landing? Are there any specific procedures that kick in when the headwind exceeds a certain number for a certain aircraft?
I don't know if aircraft type matters much in answering this question, though I realize that small aircraft could potentially encounter a headwind that is faster than the recommended landing speed, thus making it difficult to reach the runway in the first place. Let's leave that situation out of the answers - assume that the wind is slower than the recommended landing speed.
I understand that landing into a headwind is the preferred option, and that tailwind landings are possible up to certain limits.
I understand that crosswind makes the landing more challenging and that there are limits to the crosswind component. Assume a direct headwind for this question.
I understand that a headwind makes ground speed slower for a given quantity of lift, but that's irrelevant to anyone but the passenger anxious to make his connection.
Assume a fairly steady wind with minimal gusting - at a level most pilots wouldn't be concerned with. (Unless, of course, gusts make a more or less significant difference based on steady-stream speed.)
Update:
This seems to be a fairly difficult question. I get the fact that, in practice, the above conditions are likely never going to be met, this is more of a theoretical question - assuming a direct headwind, assuming near zero gusting, assuming "significant" differences in headwind (in relation to the aircraft's landing speed - a C152 with one on board will land slower than a fully loaded A380). These are combinations that will probably never happen outside a wind tunnel, but I'm sure someone, somewhere has thought about them.

If I have a 50kt headwind, and my landing speed is 65kts, that means I touch down with 15kts of ground speed. It also means I'll be flying the airplane to the hangar...

At least you'll still be moving forward, @RonBeyer, instead of backwards!

Sounds funny, but I've actually flown an airplane backwards in a strong headwind (at altitude practicing slow flight). Controller was pretty confused about the speed readout.

The main effect is in the amount of energy which must be dissipated to bring the aircraft to a stop. As Energy is proportional to velocity squared, reduction in actual ground speed at touchdown has a substantial effect on this, reducing landing roll distances significantly.

In practice no one flies in very large headwinds like 30/50 knots ( at least for small airplanes ). Winds tend to be gusty when high and 50 can go down to 10 in a second and you will stall the plane and fall to the ground.

I do not understand your update to your question.  What are you looking for that is not covered in the two answers already given?

Are there any specific advantages or drawbacks to a stronger headwind when landing?

A strong headwind right down the runway can appreciably shorten your landing roll, if you want to do that. If you want to clear the runway as soon as conveniently possible, you might be able to take a taxiway closer to touchdown than you otherwise would.
Or, let's say the taxiway that is closest to your gate is 3/4 or more down the runway, so what you might do in a strong headwind would be to not use reverse, or just go into reverse but not bring the power up. What you don't want to do, at least at busy airports, is tie up the runway at a taxi speed.
Some have contended that a strong headwind, with the resulting lower groundspeed, makes for a slightly smoother landing because the tires don't have to spin up as much. I don't have an opinion on that.
A strong headwind can mean that brake temperatures on arrival at the gate will be lower than otherwise, both because of less brake usage on rollout and better ventilation of the brakes while taxiing.

Are there any specific procedures that kick in when the headwind exceeds a certain number for a certain aircraft?

I've never heard of such specifically for a headwind, or at least I don't remember ever hearing of such.

This is from the viewpoint of a learner/novice pilot where landings are still a challenge.

Are there any specific advantages or drawbacks to a stronger headwind when landing?

Consistent and even headwind landings are lower risk and easier to do than no wind at all. Damage to aircraft and their passengers is caused by the difference in velocity between the aircraft and the ground. When landing in a strong even headwind you still have the same vertical descent speed but a dramatically reduced horizontal speed along the runway. This means that if anything goes wrong (plane not perfectly lined along the runway, landing a bit heavily etc) the ability to retain control and transition to running along the runway on your wheels is much easier. In the event of a loss of control and coming off the runway and hitting anything or indeed hitting an unexpected object on the runway the relative velocity of your plane to the stationary object involved is much less meaning less energy in the crash, less damage to everything involved and better chances of personal survival.
Also as you often are trying to keep to the same rough angle of descent you will often come in with a bit more power on which when the prop is in front of the wings increases airflow over the control surfaces making the aircraft a bit more responsive.
As a learner pilot in a little Cessna 152 (landing at about 60 knots) I greatly enjoyed a few days I got doing circuits (landing practice) with a nice strong steady 20 knot headwind as it allowed me to worry less about the wheels being perfectly ready to transition to trundling along the runway. 40 knots relative to the ground feels really slow and with a long runway that is going past slowly you have more time to focus in on getting the flare nice and accurate and touching down gently.
I did put the caveat on "consistent and even headwinds" as the reality is that stronger winds are usually more variable than weaker winds. It is not as easy if the headwind is varying by +-10 knots which is a lot more likely when the average is windspeed is 20 knots than when the average windspeed is 5 knots.

Are there any specific procedures that kick in when the headwind exceeds a certain number for a certain aircraft?

As implied above when the windspeed approaches the landing speed of the aircraft you need to apply power on descent to maintain an appropriate angle of approach. Otherwise you end up descending vertically onto the threshold.

Are there any specific advantages or drawbacks to a stronger headwind
when landing?

As for drawbacks, a stronger headwind generally means a stronger headwind gradient.  This means that for a given power setting and airspeed, the aircraft will tend to have a higher sink rate near the ground.1  In a worst-case scenario, the aircraft may simply not have enough airspeed for a flare that adequately arrests this sink rate-- one solution is to maintain extra airspeed throughout the last part of the final approach, and then bleed it off very near the ground, but preferably not in an abrupt "flare" maneuver (read on for why not).
A strong wind gradient also means that any accidental ballooning upwards due to an excessive flare2 will tend to be more pronounced, after which the airplane will tend to plummet downwards due to the effect described above.
I fly radio-controlled model airplanes as well as "real" airplanes, and those lightweight aircraft (little "foamies") are very prone to wind gradient effects.  In light to moderate winds, I usually have smoother landings when landing downwind than upwind, due to the effects described above.  (I don't land "real" airplanes downwind.)
Hang glider pilots who "flare"3 to reduce groundspeed to land on safely their feet are also very familiar with these effects-- and generally reserve a strong "flare" only for no-wind or light-wind situations, to avoid this "ballooning" effect-- though the dynamics are complicated by the fact that there's no need to flare hard enough to greatly alter the groundspeed when landing into a strong headwind anyway, since you generally don't want to touch down with a negative groundspeed, i.e. going backwards.
Footnotes:

Assuming that we're talking about a power setting and airspeed that would result in a positive sink rate (negative vertical speed) in still air, i.e. in the absence of any wind gradient.

Also pertains to an accidental upward "bounce" into the air after wheels contact the ground.

The type of "flare" used in a light-wind or no-wind landing with a hang glider is very different from what we use in airplanes to simply arrest the sink rate-- it is a dynamic maneuver that uses the inertia of the pilot's body to rapidly put the wing at an extremely high (stalled) angle-of-attack that results in a rapid loss of airspeed and groundspeed.  Improperly timed or improperly executed, it can cause the glider to climb slightly-- or more than slightly.

+1 for wind gradient. Definitely true for sailplanes. I've been taught to add half of the wind speed to the usual landing speed.

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Why does Class D overlap class B around SeaTac? ...and other Seattle Bravo airspace strangeness
I noticed KBFI's class delta overlaps SeaTac's bravo by 500 to 700 feet. Does anyone know which tower the overlapping airspace belongs to? Does it classify as Bravo airspace and do you need to hear the magic "cleared into the bravo" phrase before entering the overlapping areas?
Bonus question: There is a tiny sliver of class D airspace surrounding the surface B of SeaTac on the western, southern and southeastern parts of the bravo that is surface. Anyone know if this belongs to SeaTac because there is no other field in the center of the area and if so, why on earth does a bravo airport also have a tiny section of class D surrounding it?

This is a great question. For those who are unfamiliar, and in areas where airspace is complex, it’s sometimes useful to look at the VFR Flyway Chart. This is the one on the back of FAA Terminal Area Charts.

The dashed blue outlines surround the crazy cutouts that are Seattle’s Class D airspaces.
Why does KBFI stick into the KSEA Class B?
Class D airports, or at least those that don’t exist entirely within more restrictive airspace, will always have a ceiling. This is because there will be sections that are not contained within the Class B (or C) areas. For those areas, we need to know how high the Class D airspace goes.
If there is Class B and Class D airspace in the same place, which is it?
This one is in the AIM, 3-1-3:

Hierarchy of Overlapping Airspace Designations
a. When overlapping airspace designations apply to the same airspace, the operating rules associated with the more restrictive airspace designation apply.

So, if you’re over the I-90 bridge at 1900 feet, you’re in KBFI Class D. If you’re at 2100 feet, you’re in KSEA Class B.
Why does KSEA have weird class D chunks?
This one is easier - because Class B doesn’t always go to the surface (only at the core), but there is still a need to filter out traffic from the immediate airport environs. Because two-way communication is required, pilots will need to talk to the tower before getting too close (~5SM) to the airport.
Another factor is that Class D centers on the airport; Class B is centered on the SEA VOR, and they are not coincident. This is so that the rings could (when it was circular) be defined with DME.

Thanks. That FAR/AIM is what I was after. In your example you cite Seattle as being Class C when it's actually B (which is obvious from the map coloring). Also I'm not sure that your explanation about the thin sliver of Delta around the SeaTac bravo is 100% correct. I've never seen any Bravo that doesn't go to the surface (do you know of an example?) and why not just have SeaTac extend the surface bravo? Thanks again.

Whoops, typo. And the B does go to the surface - just not for a 5 mile radius. Seattle has pretty unique topography and it also has been updated to use the new non-round B design.

Oh, and because the B is centered on the VOR while the D is centered on the airport.

Your image has perished. :( Any chance you still have it lying around and can replace it?

The SeaTac Class D space is below the SeaTac Bravo.
Its purpose is basically that if you are getting close to the Bravo, SeaTac wants to be in communication with you, and know that you're aware of your proximity to the Bravo. But they probably won't issue you a Transponder Code or give you vectors (unless traffic is near by).
Its just gives them confidence that you're aware of where you are and what you're doing before you end up in front a 767 on short-final. :)

Bonus question: There is a tiny sliver of class D airspace surrounding
  the surface B of SeaTac on the western, southern and southeastern
  parts of the bravo that is surface. Anyone know if this belongs to
  SeaTac because there is no other field in the center of the area and
  if so, why on earth does a bravo airport also have a tiny section of
  class D surrounding it?

In the FAA's "Airspace Designations and Reporting Points" document, this airspace is specifically described as "Class D airspace... extending upward from
the surface and designated as an extension to a Class B surface area."  (page D-164).  It is the only example of this kind of airspace in the entire US national airspace system.
There is also one single example of "Class D airspace... extending upward from
the surface and designated as an extension to a Class C surface area."  It is at Biggs Army Air Field at El Paso, TX (KBIF) (page D-165).
The purpose of the Class D "extension" at SeaTac is to require approaching pilots to communicate with the control tower, without imposing the requirement for an actual clearance to enter the airspace, and without imposing the other requirements associated with Class B airspace.  Yes, it does "belong" to SeaTac.
The Class D "extension" at KBIF is more of a "hybrid" situation-- KBIF has its own control tower but is actually located in the Class C surface area of El Paso International (KELP).  The Class D "extension" is intended to facilitate operations at KBIF, and abuts the Class C surface area of KELP.
On a related note, Los Angeles International Airport (KLAX) has Class D airspace associated with it that appears to be functionally equivalent to the Class D airspace "extensions" at SeaTac, but it is not described as an "extension" in the "Airspace Designations and Reporting Points" document.  Instead, it is described on pages D-148 through D-149, in the main section on Class D airspace, section 5000, whose description (page D-1) includes the phrase "Each Class D airspace area designated for an airport in this subpart contains at least one primary airport around which the airspace is designated."  Each piece of Class D airspace at KLAX is described as part of a ring of Class D airspace that completely encircles KLAX.  There are two such rings in total, one extending to 2700' AGL and one extending to 2500' AGL.  Most of the Class D airspace in each ring is invisible and non-functional because it coincides with Class B airspace, leaving just two protruding "tabs" of functional Class D airspace that serve essentially the same purpose as the Class D "extensions" at SeaTac.
Page number references are to the FAA's "Airspace Designations and Reporting Points" document, current edition Order JO 7400.11D.

@StephenS - see edited answer--

If any other cases like LAX exist, it would be useful to add them to this answer.

The second-to-last paragraph answer of this answer could serve as the foundation for both a question, and an answer, specific to LAX.

A related ASE question could ask WHY the Class D airspace at LAX has been designated in a different manner than the Class D airspace at SeaTac.  It may, or may not, have something to do with how far the D areas project from the inner Class B surface area; at LAX the distance appears to be very short.

The LAX Class D airspace is discussed here https://aviation.stackexchange.com/questions/50109/when-is-this-class-d-airspace-not-class-d-airspace and here https://aviation.stackexchange.com/questions/84200/does-this-chart-show-a-heliport-in-class-d-airspace-without-a-control-tower

@757toga -- re " Notice the 2700 ft north and 2500 ft south sections of the KLAX Class D are not in the Class B surface area "-- not sure I follow your point here.  The same could be said at SeaTac or Biggs AAF-- the Class D bits (which happen to be called "extensions" in those cases) are not in the associated Class B or Class C "surface areas".  Right? For example, the Class C floor is 5200' in the latter case.

PS if you read the airspace description on the pages I cited for KLAX Class D, you'll notice that large parts of it are in the Class B surface area-- but those parts are not charted, and have no functional effect, because the Class B rules and regs automatically apply there, similar to what was noted in relation to SEATAC in egid's answer (see the AIM reference).

@757toga-- Re " Part of the SEATAC Class D "extension" is inside a location where the floor of the Class B is the surface. " -- would you mind explaining exactly which part you are speaking of?  Are you sure the description on page D-164 of the resource cited in my answer backs you up on this?  (See the last sentence, "excluding that airspace within the Seattle, WA, Class B airspace area.")  Is it possible that what you are referencing is actually described on pages D-70-D71 and is actually associated with Boeing field not SEATAC-- but yet note again that sentence "exclding that a/spce w/i..."

Let us continue this discussion in chat.

(Additional comments now posted in chat, thanx)

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text:
How can a hang glider take off from flat ground?
There are a number of different ways of taking off with a powerless hang glider, the most commonly used being either running down a hill or jumping off a cliff/platform. This is how I learned to hang glide and is the standard way of getting airborne for most hang gliders.
However, I recently moved to Houston, Texas which is extremely flat. As far as I can tell, there isn't a single hill tall enough to take off from within a 100 mile radius of where I live.
How can I safely get airborne when I am on flat ground?

It appears that towing (both air and ground/winch) are accepted methods, but very likely come with large safety risks which would need to be addressed by a trusted professional. Link here: Hang Glider Launch Methods

Here in the midwest, winch launches and towing with ultralights are common.  Car tow is also used some places.  I only did some quick Googling, but it looks like the Houston has at least one hang gliding club that offers training.

Trebuchet, definitely: https://en.wikipedia.org/wiki/Trebuchet

The major ways:

Run really fast into the wind at minimum angle of attack and then pull up a bit, and you'll get a little lift that you might then use to find a thermal. This is very location-dependent, and you're unlikely to fly very long at a stretch.
Get a tow from a car or a powered plane like a sailplane would. A car limits your altitude, but is cheap and easy to set up; arranging a pull pilot is a bigger deal but can get you hundreds of feet in the air.
Tether yourself to the ground with a long rope, push the control bar forward and rise up in the air like a kite. Then, cut yourself loose, pull the control bar into your stomach and cut around in a spiral to gain forward airspeed before leveling off.

Most hang gliding I've seen done has involved launching from a tall hill or cliff, then looking for a ridge or thermal updraft nearby to keep you in the sky.

If you were an avid hang glider pilot at the time you posted your question, you no doubt have long since connected into the local hang gliding community and found the answers to your questions, and have spent many joyful hours soaring under the beautiful cumulus clouds over the Texas flatlands.
Aerotow (towing up behind a slow-flying ultralight or lightsport aircraft) is one way to launch hang gliders from flat ground.  A typical tow will be to around 2500' or 3000' up.  You'll start with the glider on a three-wheeled dolly so your own legs don't have to be involved.  If you've never done it before you'll get training in a "tandem" hang glider, flying with an instructor.  There is an active hang gliding business very near Houston that offers aerotow launches, and I hope you have made contact with them since you posted your question.  (Google "Cowboy Up Hang Gliding".)
Another method is called "platform launch".  The glider is placed on a platform on the bed of a pick-up truck.  The truck has a reel of line that pays out under constant pressure, regulated by a disk brake.  The truck starts driving down a long little-used road or on a dry lakebed or an airport runway.  When sufficient airspeed is reached to ensure a strong positive launch with a strong vertical climb rate, the  pilot pulls a trigger to release the glider from the truck and the line starts spooling out.  I've gained as much as 4000' of altitude during this type of launch, using a reel of lightweight spectra line, and a 3-mile-long road.
This type of launch is a very technical procedure and unless you are a willing to take high risks as a test pilot, you must use the exact equipment that other experienced flyers have carefully developed over the years.  You must get training in the techniques, use a  detailed checklist, etc.  I'm not sure if this is currently being done anywhere near you.  Aerotow would be the easier route in most cases.
Yet another method uses a stationary winch that reels the line in.  Often such a winch is made from a motor scooter, mounted on a trailer, with the back wheel replaced by a drum (reel).  This method can be good for beginner training via ground-skimming flights, but also can be used to launch to several thousand feet of altitude.
It is also possible to tow a hang glider from a car or truck using a fixed length of rope.  This is not as simple as it sounds-- for example a line tension gauge is essential.  Read on for more safety tips.
To safely tow a hang glider, you must avail yourself of community knowledge and training.  With many methods, it is vital to have some way to actively control the towline tension, and with all methods, a weak link of appropriate strength is essential.  Reliable release mechanisms that route the towline's pull to the appropriate points on the pilot's harness (and sometimes to a point on the glider as well) are also essential.  Simply tying one end of a strong rope to the glider and the other end to a fixed point on the ground and trying to launch yourself like a kite on a windy day would be extremely dangerous.  Several fatalities occurred this way back in the early days of hang gliding, before the community gained expertise in safer methods, and also developed a strong culture of safety-consciousness.  A phenomenon called a "lockout" is waiting to bite you if things go wrong.  If you've ever seen a tethered kite dive into the ground at high speed with the string under high tension, you'll have some idea of what this might be.
Generally speaking, hang gliders have less control authority than sailplanes, especially in the roll axis, so towing hang gliders is more complicated than towing sailplanes, and more care is needed to maximize the odds of a safe outcome.  However, if due caution is taken, towing can be a very effective way to get a hang glider up into the sky.
There are also motor units available for hang gliders.  The modern method is to incorporate the motor into the pilot's harness, not to attach it to the glider itself.  Using your own feet as landing gear while launching with a motor can be somewhat challenging and also dilutes the joy of pure soaring flight; you'll probably have a more enjoyable experience if you go the aerotow route.  But google "mosquito harness" or "hang glider powered harness" for more on the motor option.
A related concept is the "light trike", where there are wheels and a seat as well as a motor, all connected to a hang glider wing, but everything is kept light enough in relation to the wing size to allow a low sink rate in unpowered flight.  The motor is used to get some altitude, and then you shut it off and have a nice soaring flight.  The control "feel" may be a little awkward compared to a pure hang glider flown from a prone position-- most hang glider pilots would prefer to let the motor fly away with the towplane at the end of the tow, as long as the soaring conditions are good and a towplane is available.
I can't answer a question about hang gliding in Texas without referencing the famous flights originating from Zapata TX on July 8 2012-- one pilot flew 764 km (475 miles), and a second pilot flew just a few miles short of that distance.  Read more about it -- and see video of their aerotow launches-- here: https://xcmag.com/news/dustin-martin-and-jonny-durand-break-hang-gliding-world-record-in-zapata-4/
And yes of course it is also possible to run into the wind as fast as you can and then increase the angle of attack of the wing to do a little hop.  In terms of peak altitude gain, this is particularly effective on a beach, where the wind gradient tends to be especially pronounced-- the wind gradient will help you out as you start to climb-- but then will work against you as you come back down.  You'll never get more than a few seconds of flight time this way, but it can be a fun thing to  do a few times after you touch down on the beach after a nice long ridge-soaring or dune-soaring flight on a windy day.

Future edits-- hide text of link urls, add more links re scooter tow, truck tow, ushpa website.

See also the comment under original question for another link, and google around for more.

You need to create lift, so (as one person mentioned) you can do this by towing - e.g. getting pulled by a car.  Then (if the air is right) you'll be able to gain enough altitude to sustain flight among prevalent currents.
You could presumably run and produce the same effect, if you had the right wind.
a small fan + motor could do similar - but now you are no longer in the range of powerless flight
you could try to fly a kite or small chute, to given you additional lift and aid powerless take-ff

Bullet one sounds like a skit idea from Johnny Knoxville.

Running won't work, because the force needs to be applied until you are high enough to be able to glide to thermals.

Also kite needs to be tethered to fly, so the last point quickly reduces to the first.

Also, a kite with sufficient strength to pull a man and his glider would be pulling you downwind, while you'd want to be taking off into the wind if at all possible. It might be possible to tack slightly, like kitesurfers do, but once you're off the ground you'd almost li8terally be trying to pull yourself along by your own bootstraps.


text:
Is excess lift or excess power needed for a climb?
As answered in this question, aircraft need excess power - not excess lift - to climb. This is plausible when the aircraft's thrust vector has a vertical component (its nose and engine points upwards), but I challenge the requirement of excess power for every case.
Please take a look at the following cart. The thrust gets delivered by a propeller at the rear and the thrust vector is always horizontal. A wing attached to a vertical beam is free to move up and down.

When the cart gets accelerated and reaches a certain speed, the lift acting onto the wing gets greater than the wing's weight, leading to a climb of the wing. Please notice that - because thrust is horizontal - the chemical energy burned goes into kinetic energy of the cart and/or heat energy (due to overcoming drag). No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift.
Did I miss something?

you're not lifting the CoG of the cart, so this is not representative of a climb situation

You did an excellent job of illustrating your question! I wish others would pose their question with such clarity.

The first thing you missed is drag: Once the wing moves through air, it will create not only lift, but also drag, and that drag will be higher when the wing accelerates up the pole. This drag increase will at least reduce the acceleration the cart receives from the engine. If the wing would not produce lift, the cart would accelerate more quickly and would settle at a higher speed.

I really like these these illustrations! Did you make them yourself? If so, what tools did you use, I need those skills as well !

I think this might be better suited for physics.stackexchange.com

@PeterKämpf: The drag coefficient of the whole machine is the same regardless where the wing is, so I don't see that the drag should increase when the wing accelerates up. If the wing would produce no lift, then the cart would indeed accelerate more quickly and reaches a higher terminal velocity (due to the absence of lift-induced drag), but this effect would also happen when the wing is not free to move.

@DeltaLima: I made them with SketchUp.

The drag of the wing changes with the square of the speed of the car, and when the wing moves up or down, it changes in addition with the third power of the angle given by the ratio of vertical to horizontal speed. A square comes from the amount of lift created, and must be multiplied by the angle again to account for its change of direction - therefore the third power.

@PeterKämpf I've added an answer to my question on lift in climb.

This answer is incorrect. If power was needed to climb, gliders would never climb. Gliders climb because of the lift provided by thermals, or other external airflow that lets them make the climb. Also, a powered aircraft can certainly climb with its engines failed.

The key to thinking clearly about this problem is to recognize that when the wing is climbing up the pole, the situation is fundamentally the same as an airplane climbing as normal, except for these changes-- 1) pitch attitude is constrained to stay horizontal, 2) this means that we have built-in down thrust relative to the direction of the flight path, and the angle of this down thrust is always exactly equal to the climb angle, and 3) the wing's angle-of-attack is constrained to decrease (get less positive or more negative) as the climb angle increases.

The point of my (locked) answer is that even if we simplify the problem by ignoring point 3, it's clear that more power is required when the cart is moving fast enough for the wing to climb, than when it is not.  We can also show that, due to the downthrust, in this specific situation lift actually must be greater than weight for the wing to climb-- unlike the case where we are flying "normally" and allowing the aircraft to pitch up as it enters a climb, so as not to force a decrease in angle-of-attack, and an increase in downthrust angle relative to the direction of the flight path.

Another interesting approach would be to hold the speed of the cart constant (which is not the same as holding the airspeed of the wing constant), and assume that the wing is initially locked in place and then we pull a trigger and allow it to start rising up the pole. Let the pole be long enough for the wing to reach a steady-state climb speed. Must thrust be greater when the wing is climbing than when it is not? Must lift be greater when the wing is climbing than when it is not?  (Do we perhaps need to know something about the Cl and Cd curves of the specific airfoil to give the answers?)

(Ctd) It might be worth creating an answer (or even a new question?) that looks at the question from this specific angle-- but keep in mind that when the wing is locked into place and therefore unable to climb, there is no longer any relationship at all between lift and weight, so this is in no way a good analogy for anything we might ever experience in an actual aircraft untethered to the ground.

The critical point that may be missed on first examination, is that this whole problem is almost identical to the case of an aircraft attempting to establish a steady-state climb while remaining in an exactly level pitch attitude.  Simplify the problem in your mind by assuming that all the components except the wing are weightless.  Since the thrust vector has no component that is perp. to the weight vector, this simplification ends up having no effect on required thrust, power, speed, or lift, at least for any steady-state situation (i.e. constant fwd spd, and const climb rate of wing.)

As the answers to your original question already explained, you do need extra lift to accelerate upwards. Once the wing is set into a vertical motion, however, lift again exactly equals weight to keep the wing at a constant vertical speed (if we neglect thrust and drag for a moment). No extra lift is needed to maintain that vertical speed. Only when you want to accelerate further up, extra lift is needed.
The increase in potential energy comes indeed from the propeller, because the lift vector of the climbing wing is tilted backwards, adding a horizontal component that needs to be compensated by extra propeller thrust.
Now let's look at your experiment in detail: I assume the wing has some mass, is rotationally locked and slides up and down that pole without friction. If you accelerate the car, at some point its speed will just be right for the wing to create exactly the lift to cancel out its own weight. At this speed the wing will be stable at any position along the pole. If it slides down a little, its angle of attack $\alpha$ will increase and create more lift, stopping the downward motion. The reverse is true for any upward motion. See below for an illustration of the principle. The cyan vector is the vector sum of the flow due to forward motion (blue) and vertical motion (red), and this is what the wing will "notice".

When the car accelerates further, the lift will increase and now become greater than the weight. The wing will accelerate upwards until its vertical speed will reduce its angle of attack by enough to reduce the vertical aerodynamic forces to exactly equal its weight. Now you have the same situation as before, but not at zero vertical speed, but at a positive vertical speed which will make sure that the wing pops out at the top of the pole unless there is some stop. When the wing hits the stop, the vertical motion ceases, the angle of attack increases and the wing will lift up not only itself, but also part of the car's weight.
Note that I now spoke of the vertical components of the aerodynamic forces, not lift. When drag is added, it will add a vertical component when the wing is in motion. Lift is defined as the sum of aerodynamic forces perpendicular to the flow direction at infinity and drag parallel to it. This cumbersome definition makes sure that local distortions in the flow field do not impact the direction of lift and drag. The direction of lift for the climbing wing will point slightly backwards and the direction of drag slightly downwards. This will add some drag component to the sum of the vertical aerodynamic forces, and lift needs to increase to compensate for this. The horizontal component of lift will now add to the drag and the forces on the pole, so more force from the propeller is needed to push the climbing wing through the air. This extra force is needed to increase the potential energy of the wing on its way up. For a descending wing, the reverse is true: Now drag will add some vertical component and lift will be slightly slower. The forward component of lift will now push against the pole, reducing the force the propeller needs to provide. The reduction in potential energy now reduces the horizontal aerodynamic forces.
An airplane is slightly different, because it is free to pitch up or down and the angle of thrust will pitch with it. This will enable the pilot to select the flight path and the amount of lift the wing creates, but again the vertical motion will make sure that any excess lift will translate into increased vertical speed and a lower angle of attack, so the excess lift vanishes. In a climb, thrust needs to be bigger than drag in order to increase the potential energy of the airplane, and now the vertical component of the tilted thrust vector will support some weight, reducing the amount of lift needed to support the weight.

If the wing doesn't rotate and the airflow is at a constant angle (i.e., in the observer's frame of reference, still air and the cart is moving along horizontal ground), how can the angle of attack change?

@DavidRicherby: Due to the wing's motion. I guess I'll update the answer with a sketch - that will be better than a wordy explanation here.

Ah, I understand now: the angle of attack is the same whenever the wing is stationary but it alters while the wing is moving up or down.

@PeterKämpf: The change of the angle of attack is actually a change in the direction of the relative wind (as seen from the perspective of the wing). Since lift is perpendicular to relative wind, there is an additional lift-induced drag onto the cart when the wing accelerates up.

There are two terminal velocities at play. When the cart accelerates, at some point it reaches its (horizontal) terminal velocity at which thrust is equal to drag. When the wing moves up, during upwards acceleration lift decreases and drag increases until lift equals drag and the wing reaches its (vertical) terminal velocity.

@Chris: Both comments are correct, and I assume you meant "…until lift equals weight and …" in the last sentence above.

@PeterKämpf: When lift is equal to weight the vertical movement of the wing would stop. So to be more precise: "until lift of the wing equals its vertical drag". Or do I understand it wrong?

@Chris: I assumed frictionless movement to simplify things. Friction will over the long term make sure that the movement stops when lift = weight, but in those ideal conditions (and for short-term effects in reality as well), only further acceleration will stop. Vertical drag would mean friction drag along the pole, right?

@PeterKämpf: With vertical drag I mean the component of the drag force acting opposite to the vertical movement of the wing, i.e. the drag force that acts downwards when the wing moves up.

As pointed out, with increased vertical velocity of the wing comes a decreased AoA --- which will decrease lift.
I would expect the vertical velocity of the wing to be proportional to the forward velocity of the cart.

@David: Not quite. With higher speed comes a higher dynamic pressure, so less angle of attack is required.

I agree with Chris. The climb speed changes the AoA, reduces up force and therefore acceleration, when up-force = weight + vertical drag.

I have a disagreement with the term "accelerate" as used in vertical motion. Established in a 500 FPM climb, there is no vertical acceleration, but there clearly is a climb. Rather energy is used to "lift" (in a non aerodynamic sense) the weight of the plane higher, in a climb, where there is no vertical acceleration.  The statement that lift equals weight to keep a constant vertical speed is not correct. If climbing, lift is greater, and if descending, lift is less.

@Peter Kampf Why the effective angle exist if it moves up or down? What happens to the triangle of wing moves upward? Will the triangle be reflection about x-axis?

@Auberron Because horizontal and vertical speed add up and produce a single speed at an angle. The speed triangle for up is indeed a mirror image of the one for down, mirrored at the x axis.

When you say,

No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift.

you're missing where the energy of the wing comes from. Lift isn't a magical power that creates potential energy out of nothing: it just turns airspeed (kinetic energy) into height (potential energy). In your example, the power invested by the propeller turns into kinetic energy of the whole cart, including the wing. That's how the energy gets from the propeller (or its fuel) into the potential energy of the wing. You need to use more thrust to drive the cart with the wing attached, than you would if you took the wing away.
There are two ways to look at the forces produced during a climb. Remember that as a wing produces more lift, it also produces more induced drag. That's why you need excess thrust, to generate the excess lift.
For a certain power setting, you can fly level at a certain speed. If you pitch up, the wings will create excess lift, but also more drag. Even though some of your thrust is acting vertically, there isn't any excess thrust, because the drag is greater. You'll slow down, the lift will decrease, and you'll stop climbing.
Instead, you can keep the aircraft level, and add more thrust. This will increase your speed, which will also increase the lift from the wings. This in turn increases the induced drag, which will eventually balance the excess thrust at a new, higher airspeed. Because you've increased the lift by doing this, you'll climb, even though your wings are level. You can only do this because you added power in the first place.
(I feel obliged to point out that you wouldn't usually climb like this: to get a better rate of climb, you'd generally add power and also pitch up, letting your airspeed decrease to the speed where the wings produce the most lift for the least drag.)

Induced drag in a steep climb is actually less than in a shallow climb, simply because lift is less (more of the net upwards force generated by thrust). By definition,  Lift and Drag are perpendicular and parallel to the flight path (relative wind),  not the earth horizontal plane. The increase in lift (and Di) s only momentary,  to accelerate to create an upwards velocity as indicated by Peter Kämpfs answer

Believe it or not, induced drag goes down when speed increases.

@PeterKämpf Because the angle of attack is decreasing, you mean? That's a point. I'd hoped to keep the explanation simpler than that, but maybe I tried to make it too simple.

Induced drag goes down when speed increases because wingtip vortices decrease at higher speed.

@DanHulme: "You need to use more thrust to drive the cart with the wing attached, than you would if you took the wing away." Of course, the reason is additional drag, which dissipates to heat. I am fully aware that this violates conservation of energy. But remember that energy conservation is a "macro principle" that gets induced by more basic principles, e.g. mechanics. You have to give mechanical reasons to show that energy conservation is in place.

In a stabilized climb (constant airspeed, constant direction of flight path through space), lift is LESS than weight.  See my answer.

The simple answer is easy to demonstrate.
Start with an aircraft TRIMMED for straight and level flight.
For example 1000 feet, 100 mph, 1500 rpm fixed pitch prop.
Lift = aircraft weight and thrust = aircraft drag.
Now increase engine rpm by 150 rpm (10% more thrust),
which increases thrust.
The aircraft will for a moment accelerate, the increased airflow
over the wing and stabilizer increases lift and the aircraft will gain altitude.
In a few seconds the system will balance once again, the airspeed will return to the  trimmed 100 mph, and the excess thrust will show up as
climb rate. The aircraft will now be slightly pitched up, but the angle of attack remains constant since it is controlled by the stabilizer trim setting, which we did not touch.
Next roll the elevator trim forward, which will lower the nose a bit.
The airspeed will increase slightly and the climb rate will reduce.
When trimmed once again to straight and level flight the aircraft
rate of climb will be 0, the airspeed will be above 100 mph.
Now the extra thrust shows up as increased speed.
To continue the example, reduce the rpm back to the original
1500 rpm. leave the trim alone.
The aircraft should now show a decent rate, at the new slightly higher airspeed.
All this was done without input from the control stick.
Anytime the pilot maneuvers  the primary flight flight controls,
there is a nearly instant trade between angle of attack, speed, lift, drag, inertia, climb rate or decent.
Jerry S.

Lift = aircraft weight is valid only in a specific scenario (and the requirement for trimmed flight does not cover it): pitch angle null and engine mounting pitch null; alternatively, thrust vector "pitch" null. In any other case, including trimmed conditions, lift != aircraft weight

@Federico, in a systemic sense, lift of the caused by engine pitch is lift. Just as body lift, tail lift (or negative lift) all sum to the body or system lift. If the aggregate lift goes up, the airplane can climb. If it becomes less than the weight of the aircraft, the aircraft descends.

In a stabilized climb (constant airspeed, constant direction of flight path through space), lift is LESS than weight.  See my answer.

This answer overlooks the fact that trim fundamentally controls angle-of-attack which is NOT the same as airspeed.  Since lift is less in a climb, for a given a-o-a we'll trim slightly SLOWER in a climb.  And same for a dive.  Excluding complexities like height of thrust line relative to CG,  downthrust angle, etc.

Interesting example.
MY INTERPRETATION OF THE QUESTION [NEW]

Here is an image from GlobalSecurity.org which I have modified to illustrate my interpretation of the question. The angle of attack of the wing is exaggerated in the original image.
My interpretation is that the question involves only vertical movement of the wing - e.g. along the vertical red lines from the lower image to the higher image. The question noted that "the thrust vector is always horizontal". And, as the image above shows, the direction of the relative wind in relation to the wing does not change when the wing moves up and down.
MY CONCLUSIONS
Yes, I believe that you could make the wing move up the pole. You would have to make sure that the wing has a positive angle of attack (AoA) since that is required to generate positive Lift. The AoA can be created by the shape of the wing or by tilting the wing back. And the engine would have to be able to push the cart fast enough to use that AoA to generate the Lift necessary overcome the Weight of the wing.
The standard equation for computing the force of Lift is: Lift (lb) = coefficient of Lift X dynamic pressure (lb/ft^2) X wing area (ft^2).

As illustrated above, a commonly used estimate is that the coefficient of Lift is 1/10 of the AoA. A commonly used estimate of maximum AoA is around 16 degrees - but that varies with each wing.
The dynamic pressure is 1/2 pv^2 where p is the air density (.00239 slugs at sea level) and v is the velocity (ft/sec). So the faster you go and the greater the angle of attack, the greater the Lift. Thus, for example, if the wing area is 6 ft^2, the cart is traveling at 30 mph (44 ft/sec), the air density is .00239 slugs and the coefficient of Lift is .25 (2.5 degree AoA), the wing would generate Lift of 3.45 lbs.
I should note that it is also possible to force the wing upwards even if you exceed maximum AoA - the same way you can move your hand upwards by sticking it out the window of a moving car and tilting it back. But that is not an example of true Lift, you are simply deflecting the wind downwards which shoves your hand upwards. The resulting Drag would be extremely high.
True Lift is the result of air pressure differentials. Within the limited AoA range, this develops enormous Lift and much less Drag. The Wright Brothers were able to fly because they understood the power of Lift and because they had created an engine with just enough power to generate the speed necessary to create enough Lift to exceed the Weight of their aircraft.
(EDIT - An interesting example of this difference can be seen with sailboats some of which are using wingsails to increase Lift.)
The title of your question might create some confusion because one could argue that you need "excess Power" to generate the speed required to create "excess Lift". In the literature, the term "excess Power" is used to describe the Power Available in excess of the Power Required for level flight at a constant speed. This excess Power can be used to either increase speed or to increase Lift at that speed.
However, I am limiting my response to the exact question you posed in the body of your discussion.
ADDITIONAL NOTE [NEW]
One problem with the vehicle design in the example is that it will be hard to stop the wing from flying off or hitting the top of the beam. Once the vehicle reaches the speed required to cause the wing to generate excess lift nd maintains that speed, there is nothing to stop the wing from continuing to rise. But I don't believe that is a problem because the purpose of the question was only to determine whether the wing would start to rise vertically.

Comments are not for extended discussion; this conversation has been moved to chat.

The above answers beautifully explain the theoretical solution to your problem, but since you haven't accepted any one of them as of now, i'd be illustrating the solution numerically.
Lets assume that your cart is moving with a constant velocity of 'v'

Then, K.E. = 1/2 (mv^2)
D = 1/2((density)(v^2)S(Cd))
and total energy E = K.E. + D*distance (Assuming frictionless interaction of surfaces everywhere)

now, Cd = Cd0 + K(Cl)^2
distance = v*t
so T.E. = 1/2(v^2)(m + (density)SVt(Cd0 + K(Cl)^2))

Here it can be seen that total energy is being used for

The kinetic energy part of the Cart
The coefficient of lift part of the Cart's wing

The coefficient of lift part is hence responsible for the energy use3d up in lifting the wing upwards and hence the whole system obeys conservation of energy

Updated simulation without vertical drag force
In this situation, for the wing only, lift in a climb is greater than the weight. The vertical force stabilises to equal the weight, but since the lift vector is tilted backwards slightly  due to the upward velocity, aerodynamic lift increases.
Peter Kämpfs answer describes what happens to the wing in this situation, but what we did not have was a quantification. I've run a real time simulation of the forces on the wing in the drawing of the OP, as a function of airspeed $V_{air}$ and vertical wing velocity $\dot{z}$. The forces on the wing are drawn below, I've taken a NACA 0012 profile with an $ \alpha_0$ of 2 degrees:

$$L = lift =  C_L \cdot \frac{1}{2} \cdot \rho \cdot {V}^2 \cdot A \tag{1}$$
$$D =  C_D \cdot \frac{1}{2} \cdot \rho \cdot {V}^2 \cdot \tag{2}A$$
For NACA 0012, $C_L$ is proportional to $\alpha$: $C_L$ = 1 at $\alpha$ = 10 degrees, hence
$$C_L = k_L \cdot \alpha \tag{3} $$
When the wing goes up, the angle of attack changes:
$$ \Delta \alpha = arctan(\frac{\dot{z}}{V_{air}}) \tag{4}$$
We now lump all the constants together: $K_L = k_L \cdot \frac{1}{2} \cdot \rho \cdot A$, $K_D = 0.01 \cdot \frac{1}{2} \cdot \rho \cdot A$  ($C_D$ for standard roughness at Re = 6 x $10^6$ = 0.01 for angles up to 4 deg)
The lift of this angle of attack is found by combining (1), (3) and (4):
$$ L = K_L \cdot (\alpha_0 - \Delta \alpha) \cdot V^2 \tag{5}$$
resulting force $F$ is divided by mass to result in wing acceleration, which is then integrated with a digital Euler integrator to yield $\dot{z}$
L and D are aligned with the free stream vector V, while the weight is always aligned with the vertical. We take the cosine of the L vector minus the sine of the D vector
$$ F_{up} = L \cdot cos(\Delta \alpha) - D \cdot sin(\Delta \alpha) \tag{6}$$
Now for:

m = 1 kg
A = 1 $m^2$
$\alpha_0 = 2 deg$
$k_L$ = 0.1

We get L = 9.81 N at $V_{air}$ = 8.949 m/s. If we then increase $V_{air}$ from 8.949 to 10.5 m/s in 1.5 second, the wing gets an initial acceleration upwards. After 2.4 seconds acceleration is zero, the wing goes up with constant velocity $\dot{z}$ = 0.1 m/s. The angle of attack has then reduced from 2 deg to 1.45 deg
Values printed for beginning of test through to 3 seconds:

There are some 2nd order effects in the response which may be from digital instability due to the large time step of the Euler integrator. Time to check this is not available at the moment.
So in the end situation, L is 9.82 N which is larger than weight in a climb due to increase in airspeed. Not by much - the lift vector is tilted backwards at a small angle, determined by the ratio of $\dot{z}$ and V which is 0.01. The total vertical force is $ L \cdot cos\alpha - D \cdot sin\alpha - W$

You start from wrong assumptions. Drag is always perpendicular to lift - it's defined this way. There cannot be a drag component aligned with lift. Next, when the wing climbs, the lift force vector tilts backward and the drag vector downward due to the change in angle of attack. Now some of the drag is in the direction of weight but still perpendicular to lift. What is missing is the horizontal force of the beam which counteracts drag.

@Peter I've thought long and hard about your comment. It makes sense to always define lift and drag in the direction of the airstream, *regardless of the direction of the airstream itself", is basically what you are saying. I agree with that, and will rework the simulation model. However, there is a lingering issue that multiple persons here have been trying to express: we accept that in a climb, power is required to overcome the increase in potential energy from the gravitational field. However, where is the power accounted for that needs to overcome aerodynamic resistance in vertical...

...direction. A helicopter taking off vertically needs increased lift to offset the vertical fuselage drag. A rocket taking off needs to provide thrust equal to (weight + drag). Why is this for fixed wing only valid in thrust direction, and not in lift direction?

Thrust is opposite to drag (roughly) but larger in a climb. So all vertical drag is compensated by thrust, and some vertical thrust remains to reduce demand for lift. Please see the drawing here and look how the force vectors compare. In a descent, thrust is smaller, leaving some vertical drag uncompensated which now reduces the required lift. Yes, in a descent drag helps to reduce lift. Thrust is in the direction of drag while lift is orthogonal to both. So thrust compensates

drag, not lift. In vertical ascent (rocket or helicopter) it is thrust, not lift, which compensates for drag (and needs to be a bit larger than weight). Both produce thrust to counteract weight, and only when the helicopter adds some forward speed, lift will be created in addition to thrust.

@PeterKämpf I've taken the situation from your answer and quantified it, the outcome does seem logical. It's a small increase, due to the small $\Delta \alpha$, nevertheless it is an increase.

Let's keep this Gedankenexperiment and a real climb apart. My comments refer to the whole aircraft, thrust included. Then lift will decrease slightly when the flight path diverges from the horizontal. Here we have no vertical thrust component and, therefore, the cosine of lift must equal weight. Now lift increases because thrust (the force that the beam exerts on the wing) is always horizontal.

@PeterKämpf well that's disappointing. There are many facets to the question of magnitude of lift in a climb, and most of them are brought about by the axis transformation from aircraft to earth. The situation in this question totally bypasses all of the confusing issues from definitions - for instance, tilt the aircraft back, decompose the weight vector into aircraft axes, and part of the weight now becomes drag.

Tilting the aircraft back will not suffice - you need to tilt the flight path. Then a weight component in drag direction is not confusing, but essential. How else should thrust increase the potential energy of the climbing aircraft? The work done by thrust on the drag-wise weight component adds to potential energy. In this experiment, where thrust is always horizontal, lift has to take over that role.

In a steady-state climb in an airplane, the basic formula for the magnitude of the Lift vector is Lift = Weight * cosine (climb angle).  For much more on this, see the vector diagrams and calculations in this related ASE answer.  As long as the Thrust line is aligned with the flight path rather than being tilted up or down, the relationship Lift = cosine (Weight) is "baked into the physics" of a steady-state climb, even when we're flying at a high angle-of-attack that yields a high lift coefficient and a high L/D ratio.  As long as Thrust is aligned with the flight path, Lift is less than Weight in a steady-state climb.
If any downthrust or upthrust is present relative to the direction of the flight path, then the situation becomes more complicated.  Downthrust makes the Lift vector larger, and upthrust makes the Lift vector smaller.  In the limit case where Drag is zero, Lift = Weight  * cosine (climb angle) + Thrust * sine (downthrust angle), where the downthrust angle is defined relative to the direction of the flight path.  (Treat upthrust as negative downthrust.)  When Drag is not zero, then this formula no longer applies.  When Drag is not zero, if there is downthrust, the Lift vector will be larger than the value given by the above formula, and if there is upthrust, the Lift vector will be smaller than the value given by the above formula.  In such cases, the exact values of the Lift and Drag and Thrust vectors may be found by a vector diagram, if the L/D ratio and climb angle and downthrust or upthrust angle are known.
If enough downthrust is present, Lift will be greater than Weight in a steady-state climb.  We can show that in the limit case where Drag is zero, whenever the downthrust angle relative to the direction of the flight path exceeds half the climb angle, Lift will exceed Weight.  When Drag is not zero, the downthrust angle that makes Lift equal Weight becomes smaller than half the climb angle.
The cart of the "propeller train" can only push horizontally forward on the wing.  So in the case of the "propeller train", as soon as the wing starts rising, the direction of the instantaneous flight path of the wing is no longer parallel to the thrust line, so downthrust exists.  In fact, the downthrust angle is always exactly equal to the instantaneous climb angle of the wing.  So whenever the wing of the "propeller train" is climbing, Lift must be greater than Weight, even in the steady-state case where acceleration is zero.

The vector diagrams immediately above show the forces in a steady-state climb for the wing of the "propeller train" on the left, and for a conventional airplane (where the downthrust angle at the given angle-of-attack happens to be zero) on the right.
For simplicity, the Lift to Drag ratio is 3:1 in all cases.  10-degree and 30-degree climb angles relative to the horizon are illustrated for both the "propeller train" and the conventional airplane.  The diagrams are drawn to the same scale-- the Weight vector is identical in each diagram.  The magnitudes of the airspeed vectors are scaled in proportion to the square root of the magnitude of the Lift and Drag vectors.
The right-hand diagram may look a bit odd with the "unattached" Drag vectors, but look closely and you'll see two closed right triangles, each comprised of Weight, Lift, and (Thrust minus Drag), with Lift = Weight * cosine (climb angle) and (Thrust minus Drag) = Weight * sine (climb angle).
(As an aside, the right-hand diagram may be easily modified for any L/D ratio by erasing or extending the right-hand ends of the Drag and Thrust vectors, keeping the value (Thrust-Drag) the same.  It's not so simple to modify the left-hand diagram for different L/D ratios--the diagram must be completely re-drawn.)
Here is a table of the forces involved, with L/D fixed at 3:1, Weight arbitrarily set at 100, and airspeed ("A") in arbitrary units.  We also have included the Power required (P).  Since Power = force * the velocity component along the direction that the force is acting, Power is computed by multiplying Thrust times Airspeed times the cosine of the downthrust angle.
(3/1 L/D)
             "Propeller Train"              "Conventional" Airplane
Climb angle  L    D    T    A    P          L     D   D-T   T    A    P
      0      100  33   33   100  3300       100  33   0     33   100  3300
     10      115  38   69   107  7260       98   33   17    50   98   4920
     30      145  48   114  120  11900      87   29   50    79   93   7300

It's clear that more Power must be applied to the wing of the "propeller train" when the wing is rising upward than when it is not, just as with the "conventional airplane".
Just for comparison, here are the forces and power requirements resulting from an L/D of 10/1 rather than 3/1-- (the vector diagrams corresponding to these cases are not reproduced here)--
(10/1 L/D)
             "Propeller Train"             "Conventional" Airplane
Climb angle  L    D    T    A    P          L     D   D-T   T    A    P
      0      100  10   10   100  1000       100  10   0     10   100  1000
     10      104  10   29   102  2890       98   9.8  17    27   99   2700
     30      125  12   75   112  7280       87   8.7  50    59   93   5460

Warning-- we shouldn't assume that the 3/1 L/D table and the 10/1 L/D table could both reflect the same aircraft or wing, at two different angles-of-attack.  If we want to use the tables in that way, we must ignore the "Airspeed" and "Power" columns.  The reason for this is that the Airspeed and Power columns are arbitrarily set for each table so that a Lift vector of 100 correlates to an Airspeed of 100.  In reality, when we put the stick or yoke forward and decrease the wing's angle-of-attack and L/D ratio, we also decrease the lift coefficient, causing a large increase in airspeed that it is not reflected in the tables.  The "Airspeed" and "Power" numbers on the tables are only meant for comparison between different climb angles at the same L/D ratio, or for comparison of the "propeller train" and "conventional airplane" cases at the same L/D ratio and climb angle, assuming that the lift coefficient is the same in each case.
Note that in reality, unless the wing of the "propeller train" is free to pivot in pitch, the angle-of-attack of the wing of the "propeller train" cannot actually stay constant regardless of climb rate-- the higher the climb angle, the more the "relative wind" comes down from above, relative to the horizon.  This decreases the airfoil's angle-of-attack.  Therefore the L/D ratio cannot actually stay constant as the climb angle increases.  Unless the wing is fixed at a very nose-high pitch attitude so that it is essentially stalled when it is not rising, higher climb angles will be correlated with lower lift coefficients, and therefore with higher airspeeds.  And as long as we are on the "front side" of the Drag curve, the lower angles-of-attack caused by higher climb angles will be correlated with lower L/D ratios.  With the "propeller train", in the steady-state condition, a decrease in L/D ratio is reflected not as a decrease in Lift, but rather as an increase in Drag, Thrust, and Lift.  This is best visualized by looking at the left-hand vector diagram and seeing what else changes when we increase the length of the Drag vector, while holding the climb angle constant.
In essence, the vector diagrams and tables above give some insights into how the downthrust angle affects the forces acting on the wing of the "propeller train", and they also clearly show that the wing of the "propeller train" demands more Thrust and Power when it is rising than when it is not, but they pertain most directly to a wing free to pivot to maintain a constant angle-of-attack and L/D ratio regardless of the fuselage pitch attitude, along the lines of the "freewing" concept (link to PDF) (better link needed).
Another interesting variation of the "propeller train" concept would be , or a spinning cylindrical wing using the "Magnus" effect (Wikipedia link) in which case the magnitude of the Lift and Drag vectors would be insensitive to the direction of the relative wind.
What insights does the "propeller train" give us about flight in a conventional airplane?  It tells us that it is possible to enter a climb without pitching up at all, so that the aircraft's pitch attitude stays the same as it was in horizontal flight, but in most cases it is extremely inefficient.  If we are initially flying very slowly, on the "back side of the Drag curve", at a very high angle-of-attack and a nose-high pitch attitude, adding power while holding the pitch attitude constant will indeed cause us to gain airspeed and start climbing.  In this part of the flight envelope, this strategy works just fine.
However if we are in anything like normal cruising flight, we'll find that adding power while holding the pitch attitude constant can only produce a very modest climb rate and climb angle.  As we enter the climb, to prevent the aircraft from pitching up to maintain the same angle-of-attack it had in horizontal flight, we'll be holding forward pressure on the control stick or yoke to decrease the angle-of-attack, or we'll re-trim the aircraft to accomplish the same thing.  Unless we're on the "back side" of the Drag curve, the decrease in angle-of-attack associated with climbing will degrade the L/D ratio, and unless our aircraft had substantial upthrust relative to the flight path when it was flying horizontally, any positive climb angle will introduce downthrust relative to the direction of the flight path.  If we're starting from normal cruising flight on the "front side" of the Drag curve, all these relationships work against us, and we'll find that it takes a very large increase in airspeed, Thrust, and Power to achieve a very modest climb rate and climb angle.
The "propeller train" thought experiment makes it clear why we normally enter a climb by allowing the aircraft to pitch up so that the angle-of-attack is not decreased by virtue of the aircraft's rising trajectory, and the thrust line is not pointed downward relative to the flight path, regardless of whether or not we actually choose to increase the angle-of-attack as we enter the climb.

No power invested by the propeller goes into potential energy of the
wing; the climb of the wing is done purely by lift. Did I miss
something?

Yes-- with the "propeller train", when the wing is rising, thrust certainly is doing work along the direction of the flight path through the airmass, which is never purely vertical.  Here's an analogous situation-- imagine a kite flying on a windy day, with the string at a 45-degree angle.  Imagine that we have a lightweight ball that can dangle from a hook that we can clip over the kite string.  The purely horizontal force from the wind will make the ball run up the kite string.  The wind is doing work that increases the potential energy of the ball, and the propeller of the "propeller train" is doing work that increases the potential energy of the wing.
Related links:
Does lift equal weight in a climb? -- includes vector diagrams
Are there any situations where having high lift but low lift to drag ratio would be beneficial? -- includes a focus on lift and drag coefficients as well as the actual force vectors, in climbing flight
Is excess lift or excess power needed for a climb? -- examines the present question in exhaustive detail, not limited to the steady-state condition of constant climb rate, and includes a modified version of problem where the wing of the "propeller train" is free to pivot to  maintain constant angle-of-attack

Let us continue this discussion in chat.

I decided to delete the other answer so as not to distract from the present one, but it is still there those w/ +10K to see-- still contains a few significant insights not included in the present answer, and is entirely accurate as far as I can see-- https://aviation.stackexchange.com/questions/15243/is-excess-lift-or-excess-power-needed-for-a-climb/57034#57034

You have been asked before to stop making excessive edits to posts. I am locking this to prevent the same abuse of the system you have previously been notified of.

The other answer (and the basic premise) are misleading. Aviation is a careful balance of just about everything, and you rarely get more of one without losing some of the other.
Consider a rocket. No wings, and thus no lift. Sea level, 0m/s to orbit on 8 minutes. Everything done with a rather ridiculous amount of power.
Now consider an airship. Also no wings, but excess lift. Goes up all by itself. The engines are purely for maneuvering, if you remove them we usually call the thing a balloon.
One of the most ridiculous things I ever heard was a flight instructor who claimed that the throttle controls altitude and the elevator controls speed. I called complete BS on the statement and asked if he found himself flying toward some cumulo-granitus would he prefer to add power or yank back rather sharply on the controls?
What he was trying to get across was that in the extremely narrow regime of straight and level flight*, adjusting power will affect speed which affects lift, and the plane will eventually stabilize at a different altitude after a power adjustment. Adjusting pitch will change your speed almost immediately, but your altitude will change as well. You could arrange a demonstration of adjusting power and pitch simultaneously and having nothing else change, but that basically proves my point. If you want to slow down (for example, landing) do you leave the throttle at full and pull the stick all the way back? Of course not. It's a very delicate balance, and it's something pilots spend a lot of time learning. Or if you work for certain asian airlines you just have some very expensive computers handle it for you.

Yes, we spend a lot of time there, but consider how precise you need to be. A fraction of a degree, uncorrected, in any axis will result in you crashing before you run out of fuel.

"The other answer" which one? There are now three other answers and there may well be more by the time you read this.

Also, -1 for the racist reference to "certain asian airlines".

Not only that, but there were at least two answers at this time this was posted.

@DavidRicherby in case you have been living under a rock recently, I'm referring to Asiana's preference for autoland rather than proper training. If your pilots can't fly a visual on a clear day then something's wrong. It's a corporate attitude, and they just happen to be located in Asia. And BTW I happen to live in Asia. I also happen to be married to an Asian.

@paul OK. But, as you say, they just happen to be located in Asia. Saying they're Asian doesn't identify them and their ethnic origin isn't relevant to their preference for autoland. Using the word "Asian" as the only description makes it look like you're talking about Asian airlines in general so it's best avoided.

@DavidRicherby That's why I said "certain asian airlines" and not "asian airlines". I wonder what your reaction would have been if I said "certain european airlines".

In "One of the most ridiculous things I ever heard was a flight instructor who claimed that the throttle controls altitude and the elevator controls speed.", you miss the distinction between transient and steady-state effects.


text:
If a beginner paraglider / paramotor wing fully stalls, how much vertical distance is needed for recovery?
In paramotoring, a beginner wing is a wing with A, B, C, D and F (brake) lines.
When a full stall is induced, a pilot may fall a long, long ways before the wing will recover, and most will need to plan for it.
On a calm day, what is the vertical recovery distance for a beginner wing?
As an example, one could use the Gin Bolero 6

It's difficult to find numbers like that.  It's going to depend on how deeply you get into the stall.  If you push it until the wing is folding back on itself, you are going to lose 50ft minimum even on an EN-A wing.  If you get off the brakes as soon as the wing starts to fold, you may only lose 10-20 ft.

"a beginner wing is a wing with A, B, C, D and F (brake) lines" this isn't necessarily true..  a beginner wing is one with good handling characteristics for beginners regardless of construction.

@JeffUK is there a wing you had in mind when you say this?

@tuskiomi Independence Cruiser3 is EN-A rated (beginner) and only has 3 risers + brakes

Do paragliders use the same wings as paramotors?  If so the title could say "paraglider/paramotor" instead of "paramotor" (my recent edit)

I read in the link that "The Bolero 6 is intended as a first paraglider, " so I assume the answer to previous comment is "yes" and will further edit accordingly; feel free to roll back if this is wrong

Any extra second in recovery adds up to 60 meters of height. For this reason there is no way to tell how much height the next stall will require.
From the manual of Gin Bolero 6:
Page28, Dynamic Stall: the sink rate in a controlled stall induced by pilot is approx 8m/s.
Page20, Spiral Dive: you need 120-150m clearance to attempt a spiral Dive.
If you took an SIV course and trained in Dynamic Stall, you will find the answer. This is as far as I can go at answering your question. Any further appreciation or direct answering would be a mistake and huge liability.

Welcome to aviation.SE! I guess I'm misunderstanding what you mean when you say a) 1 second of recovery is up to 60m, and b) the sink rate in a controlled stall is 8m/s. That's a huge difference, can you give some more details on what it means?

The terminal velocity in free fall is roughly 60m/s. So at terminal speed every second is worth up to roughly 60m of height. As for the figure of 8m/s, that is quoted from the manual of the paraglider suggested in the question. It says there that a stall induced by the pilot leads to a sink rate of 8m/s.

The stall is not something the wing would recover without active piloting. Please check the dynamic stall portion in the manual of the wing you have specified. If the pilot doesn’t control the wing, the stall becomes irrecoverable. For this reason is not like if the pilot induced the stall then released the brakes the wing would recover within that height and time. All the pilots presenting collapse testings are doing those well within the self-recovery envelope of the wing and in ideal flight conditions. The stall and collapse accidents happen outside the wing self-recovery.


text:
What is this airstrip in Mali?
While randomly browsing on Google Earth and Bing Maps I found an airstrip in Mali.
Normally you assume a small village/mining site in the vicinity of the airport but other than the ruins of a supposed old settlement nearby I could not find anything.
Does anyone know the name of the airfield or any info why it is there? On Google Maps there is an image attached with an airplane (Beech King Air maybe) but not convinced it was taken there.

it looks like there are some ruins of an abandoned walled town nearby, as well as an older abandoned airstrip. Maybe something used by archeological expeditions?

I tried to search on other geographical tools but without luck for now. No labels, no insightful photos geolocated nearby. Now I'm more and more curious.

I can't provide much concrete information on the airstrip itself, but the King Air photo is one operated by Sahel Aviation Services within Mali. I know several individuals who worked for SAS and they have done a lot of work for the mineral industries in Mali.

This ENAGEO presentation also references the airstrip and its use in supporting the exploration process.

@JWalters: Nice find! I've added the information on the ENAGEO presentation to my answer.

I'm surprised that the "ancient aliens" hypothesis has received no consideration here.

This seems to be the "Assedreme airport and seismic camp", according to Openstreetmap. Aviation Africa has a picture.
This is most likely a now disused airstrip for a base camp for seismic oil and gas exploration. Seismic exploration is a technique for searching for oil and gas deposits by measuring the reflection of artificial seismic waves (simple introduction at oilandgaslawyerblog.com).
The airstrip seems to have been built in 1985, expanded in 2008, and then abandoned. This fits pretty well with an article describing exploration in Mali, which states that lots of seismic exploration was carried out in Mali in the 70's and 80's, with renewed interest in the 2000's.
As J Walters pointed out in a comment Algerian exploration company ENAGEO mentions using this airstrip for its operations in a presentation from 2009.

Thanks Jake, really good find and really interesting why they need to do seismic "research" in the middle of the desert what is more it looks it was worth "building" an airstrip there so very likely they use it frequently eventhough not much information is available on the internet

@BelaVizer: I've added some more information on the purpose of the airstrip. It's most likely for oil and gas exploration, not for scientific seismology. The strip seems to be disused by now.

How in Zeus's abandoned outhouse did you ever find that!?!?!! I mean, I'm as big a map and imagery guy as the next, and lost 80% of my productivity the month Google Earth was publicly released. Heck, I was an all source intelligence analyst for a while, and scoured me some raw imagery on occasion. But in every case, there was something to look at. That looks like a close-up of a bowl of Cream of Wheat with a hair dropped on top! What ever compelled you to search in that empty wasteland for what looks like a cigarette butt? For the record, not all of Mali is a wasteland, but that part is!

Annnd I just clicked through to the Bing Maps page in the OP. I can tell you without a doubt that airfield is periodically used by the French military as a Forward Arming & Refueling Point (FARP) as required to support operational rotary-wing assets. I've also seen fixed wing use it occasionally, but at the time I observed the activity identifying their intent was not a priority.

@BigNutz: Haha, yeah, I was also surprised that someone would come across this airstrip while randomly scrolling through Google Earth. Out of interest, how do you know that this airstrip is used as a FARP? Have you been to the airstrip in person? From the images, the strip looked pretty abandoned to me.

@Jake We were looking at the airfield and surrounding area for some reason or another, I can't remember exactly why. We had a system which enabled us to only see those raw overhead imagery renderings where the AI detected a change which met a threshold for flagging. Back when the French were very active in Mali, that airfield was flagged a few times a month. I don't remember how often we assessed the activity as a FARP, but it was a couple times a month.